By central limit theorem, we know that for independent and identically distributed random variables $X_1,X_2,...,X_n$, as $n\rightarrow\infty$, that:

$$\frac{\sqrt{n}(\bar X-\mu)}{\sigma}\rightarrow N(0,1)$$

Where, in this context, $\rightarrow$ represents convergence in distribution, $\bar X=\frac{\sum_{i=1}^n X_i}{n}$, $\mu$ is the mean of the $X_i$'s, and $\sigma$ is the standard deviation of the $X_i$'s.

However, what is the the limiting distribution of the following quantity, for $\alpha\in\mathbb{R}_{>0}$ and $\beta\in\mathbb{R}$:

$$\frac{\alpha\sqrt{n}(\bar X-\beta\mu)}{\sigma}$$

Similarly, what is the limiting distribution of:

$$\frac{\alpha(\bar X-\beta\mu)}{\sigma}$$

Is it possible to determine a limiting distribution for either of these quantities for all values $\alpha$ and $\beta$? Remember, the underlying $X_i$'s are not necessarily normally distributed; all we know is that they are i.i.d. If not, what additional assumptions need to be made to determine these limiting distributions?


Let us start from the second expression. Note that $\bar{X}_n$ converges in probability to $\mu$ (WLLN), and all the rest are constants, hence by the continuous mapping theorem you have that $$ \sigma^{-1}(\bar{X}_n - \mu \beta) = g(\bar{X}_n) \xrightarrow{p} g(\mu) = \sigma^{-1}(\mu - \mu \beta). $$ For the first expression is slightly more subtle. $\alpha$ doesn't bother us, however let us look at $\beta$. If $\beta = 1$ then you limit is $\alpha N(0,1)$ (by the continuous mapping, regular calculations - whatever). Now, let $\beta \neq 1$ (and $\mu \neq 0$), thus $\pm \mu$ in the brackets you get $$ \frac{\alpha \sqrt{n}}{\sigma}( \bar{X}_n - \mu + \mu(1-\beta) )= \frac{\alpha \sqrt{n}}{\sigma}( \bar{X}_n - \mu) + \frac{\alpha \sqrt{n}}{\sigma}\mu(1-\beta) \xrightarrow{D}\alpha N(0,1) \pm \infty = \pm \infty. $$ Namely, if $\beta < 1$ then it goes to $\infty$, if $\beta >1$ then to $-\infty$.

  • $\begingroup$ In your first statment, you say that $\sigma^{-1}(\bar{X}_n - \mu \beta) = g(\bar{X}_n) \xrightarrow{p} g(\mu)$. However, $g(\mu)$ is just a constant, is it not? Should it not converge to a random variable? I'm specifically interested in the distribution of what it's converging to. $\endgroup$ – jippyjoe4 Feb 16 '18 at 3:33
  • $\begingroup$ @jippyjoe4 There is no reason for $g(\bar{X}_n)$ to converge to a non degenerate r.v. $\bar{X}_n$ converges a.s. to $\mu$ and $g()$ is just some continuous transformations, hence $g(\bar{X}_n)$ converges a.s. to $g(\mu)$. Intuitively, In the second expression the $\sqrt{n}$ is the factor that doesn't let $\bar{X}_n - \mu$ to go to $0$ as it "pools" it further as $\bar{X}_n$ tends to $\mu$, while in the first expression there is no such factor. $\endgroup$ – V. Vancak Feb 16 '18 at 10:44
  • $\begingroup$ Okay, I think that I understand now. Thank you. $\endgroup$ – jippyjoe4 Feb 16 '18 at 17:56

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